Help understanding the proof: $\forall k \in \mathbb N [(a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1}]$ In Spivak's Calculus, one is asked to prove that $(a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1}$
I will write his proof (from the solution manual) verbatim, but then I have some pedantic questions at the end, which are the primary purpose of this post.

The only extra piece of information Spivak provides is:
$\text{P1}$. $a+(b+c) = (a+b)+c \quad$ Associative Law for Addition
and the following definition:
$a_1+a_2+a_3+...+a_{n-2}+a_{n-1}+a_n:= a_1+(a_2+(a_3+...+(a_{n-2}+(a_{n-1}+(a_n)))...)$
However, it seems to me that he should have also mentioned that his bracket symbols $[$ and $]$ are presumably meant to represent different layers of the parentheses symbols $($ and $)$.

Base Case:
$a_1+a_2 = a_1 + a_2 $
Assume:
$(a_1+...+a_k) + a_{k+1}=a_1+...+a_k+a_{k+1}$
Prove the $k+1$ case: $(a_1+...+a_k + a_{k+1})+a_{k+2}=a_1+...+a_k+a_{k+1}+a_{k+2}$
\begin{align}
\text{ If the equation holds for } k \text{ then } \\
(a_1+...+a_k + a_{k+1})+a_{k+2}&=[(a_1+...+a_k)+a_{k+1}]+a_{k+2} \\
& \quad \quad \text{ by P1} \\
&=(a_1+...+a_k)+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{since the equation holds for } k \\
&=a_1+...+a_k+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{by the definition of } a_1+...+a_{k+2}\\
&=a_1+...+a_{k+2} \\
\end{align}

First Question
I do not understand Spivak's base case. Shouldn't his base case start as $(a_1)+a_2$? It seems to me the only way of going beyond this sentence is to first make the definition that $(x):=x$. However, in making this definition, doesn't the base case already solve everything? Because any finite expression of $k$ terms can be lumped together, called $x$, and then you'd have $(x)+a_{k+1}=x+a_{k+1}$...which is the whole point of this proof.
Edit: A work around for this seems to be to dictate that this proof only operate on $k\geq 2$. Because if so:
We could say $(a_1+a_2) +a_3=a_1+a_2+a_3$ because by definition,  $a_1+a_2+a_3=a_1+(a_2+a_3)$ and then we can invoke the associative property to PROVE .$(a_1+a_2) +a_3=a_1+a_2+a_3$.

Second Question
The offending transition that has confused me is when Spivak writes:
\begin{align}
&=(a_1+...+a_k)+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{since the equation holds for } k \\
&=a_1+...+a_k+(a_{k+1}+a_{k+2}) \\
\end{align}
Now, I imagine what Spivak is trying to convey is that we can think of $a_{k+1}+a_{k+2}$ as being a single term...say $\alpha$...and that our assumption in our Assume step can be reframed as $(a_1+...+a_k) + \alpha=a_1+...+a_k+\alpha$.
From what I've learned in set theory, $a$ is actually a function with a domain of some subset in $\mathbb N$ and, in this context, a range that is a subset of $\mathbb R$. Given this, I do not see why at all our initial assumption case of $(a_1+...+a_k) + a_{k+1}=a_1+...+a_k+a_{k+1}$ can be applied to $(a_1+...+a_k)+(a_{k+1}+a_{k+2})$. In general, the quantity $(a_{k+1}+a_{k+2}) \neq a_{k+1}$.
So how exactly is this step actually being applied?
Edit: The only thing I can think to do here is invoke ordinals; I will explain. Originally, the formal definition of the proof I thought was: $\forall k \in \mathbb N [(a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1}]$. However, to get around the issue of $(a_{k+1}+a_{k+2}) \neq a_{k+1}$, it seems like I should instead write the proof as:
$\forall k \in \mathbb N \Big [ \forall a \Big( [\text{func}(a) \land \text{dom}(a) \in \omega \land \text{ran}(a) \subseteq \mathbb R] \rightarrow (a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1} \Big) \Big]$
In doing this, when I arrive at my assumption case, I assume for some $k$ that:
$\forall a \Big( [\text{func}(a) \land \text{dom}(a) \in \omega \land \text{ran}(a) \subseteq \mathbb R] \rightarrow (a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1} \Big)$
WITH THIS, when we arrive at the situation of:
\begin{align}
&=(a_1+...+a_k)+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{since the equation holds for } k \\
&=a_1+...+a_k+(a_{k+1}+a_{k+2}) \\
\end{align}
we can use our updated formula to tackle this. Specifically, because $\forall a \Big( [\text{func}(a) \land \text{dom}(a) \in \omega \land \text{ran}(a) \subseteq \mathbb R] \rightarrow (a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1} \Big)$, there definitely exists one function, call it $a'$, that has the first $k$ terms in common with $a$ but, as its $k+1$ term, has $a_{k+1} + a_{k+2}$. That is to say: $a'_{k+1}=a_{k+1} + a_{k+2}$. Then we can absolutely apply our assumption to produce: $a_1+...+a_k+(a_{k+1}+a_{k+2})$
 A: Your question fails to mention an important part of the statement of the problem: Spivak defines $a_1+a_2+\cdots+a_n$ as$$a_1+\bigl(a_2+\bigl(+a_3+\cdots+\bigl(a_{n-2}+(a_{n-1}+a_n)\bigr)\bigr)\cdots\bigr).$$First question: If $k=2$, this simply means $a_1+a_2$. So, the case $k=2$ is simply to prove that $a_1+a_2=a_1+a_2$, which is trivial. And there is no need to write $(a_1)+a_2$ since the goal of the parentheses is to tell us the order by which the sum should be done and there is no sum inside $(a_1)$.Second question: The equality$$(a_1+\cdots+a_k)+(a_{k+1}+a_{k+2})=a_1+\cdots+a_k+(a_{k+1}+a_{k+2})$$holds because of the reason that you mentioned: Spivak, at this point, is usin the induction hypothesis and he is dealing with $a_{k+1}+a_{k+2}$ as a single number. And he is doing this because it is a single number. There is nothing wrong with that.
A: First, the proof you've written contains some incorrect justifications:

$$ \begin{align}
\text{ If the equation holds for } k \text{ then } \\
(a_1+...+a_k + a_{k+1})+a_{k+2}&=[(a_1+...+a_k)+a_{k+1}]+a_{k+2} \\
& \quad \quad \text{ by P1} \\
&=(a_1+...+a_k)+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{since the equation holds for } k \\
&=a_1+...+a_k+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{by the definition of } a_1+...+a_{k+2}\\
&=a_1+...+a_{k+2} \\
\end{align}$$

Here is a corrected version:
$$\begin{align}
\text{ If the equation holds for } k \text{ then } \\
(a_1+...+a_k + a_{k+1})+a_{k+2}&=[(a_1+...+a_k)+a_{k+1}]+a_{k+2} \\
&\quad \quad \text{since the equation holds for } k \\
&=(a_1+...+a_k)+(a_{k+1}+a_{k+2}) \\
& \quad \quad \text{ by P1} \\
&=a_1+...+a_k+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{since the equation holds for } k \\
&=a_1+...+a_{k+2} \\
&\quad \quad \text{by the definition of } a_1+...+a_{k+2}\\
\end{align}$$
Spivak defines addition as an operation performed on a pair of numbers. If $a$ and $b$ are numbers, $a + b$ is a single number formed by the sum. It's probably best to put aside for now everything you know about set theory and functions. Spivak's talking only about the addition of pairs of (real) numbers.
If we have a sum involving more than two numbers, we can use brackets to specify the order in which terms are to be added together, but the end result of a sum is always a single number. $b + c$ is a single number. $a + (b + c)$ is a single number.
I'm not sure which edition of the book you're using, but in the 3rd edition version of this exercise, Spivak never uses brackets around individual terms. Expressions like $(a_n)$ don't clarify anything, since addition works only on pairs of numbers.
In the inductive argument, you can prove base cases up to n = 3 (via P1). I did this because the theorem doesn't really mean anything for less than 3 terms. (Complete induction works also.)
For the $k$-case you have:
$$(a_1+...+a_k) + a_{k+1}=a_1+...+a_k+a_{k+1}$$
The statement says the sum of $k$ terms (bracketed as specified by Spivak's convention) added to a single term produces the number equal to the sum of all $k+1$ terms, bracketed as specified.
If the last term $a_{k+1}$ is itself a sum of several terms, that's ok, because the end result of those sums is a single number: $a_{k+1}$.
It's easy to mistake indexes for the number of terms here.
Sometimes it can be helpful to step through a specific case to see how the inductive step works. Just for overkill, let's do that here for the $n = 5$ case.
Assuming it's true for 4 terms, lets prove the 5 term case. Let's show
$$(a_1 + (a_2 + (a_3 + a_4))) + a_5 = a_1 + (a_2 + (a_3  + (a_4 + a_5)))$$
Now we know from the $n = 4$ case that
$$a_1 + (a_2 + (a_3 + a_4)) = (a_1 + (a_2 + a_3)) + a_4$$
(By the $n = 4$ case assumption, we can "pop" out the innermost term $a_4$)
Thus
$$(a_1 + (a_2 + (a_3 + a_4))) + a_5 = ((a_1 + (a_2 + a_3)) + a_4) + a_5$$
$$= (a_1 + (a_2 + a_3)) + (a_4 + a_5)  \text{ (by P1)}$$
$$= a_1 + (a_2 + (a_3 + (a_4 + a_5))) \text{ (again by the $n=4$ case)}$$
Above, we've "popped back in" the last term, which is allowed because of the $n=4$ case assumption. $a_4 + a_5$ is just a number.
Finally, be warned: the last step in this exercise (not mentioned here) is kinda hairy.
